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Let $A$ be a discrete valuation ring with perfect residue field $k$ of characteristic $p$ and field of fractions $K$ of characteristic $0$. Let $G$ and $H$ be two finite groups and assume that $K$ is sufficiently large, so that the irreducible characters of both groups form a $K$-basis of the respective space of class functions with values in $K$.

If $G$ and $H$ have the same character table over $K$, i.e. if we have a bijection between the conjugacy classes of $G$ and $H$ and a bijection between the irreducible characters over $K$ under which the character table is preserved, do $G$ and $H$ then also have the same Brauer character table?

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  • $\begingroup$ Do you mean Brauer character table of irreducible representations with values in $k$? $\endgroup$
    – LeechLattice
    Sep 29, 2020 at 3:35
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    $\begingroup$ @Leech lattice: The Brauer character table customarily means that its entries are sum of complex $p^{\prime}$-roots of unity lifted (using a suitable bijection) from roots of unity in $k$. The entries are not actually elements of $k$. $\endgroup$
    – Geoff Robinson
    Oct 7, 2020 at 8:52

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